The pV diagram in (Figure 1) shows a process abc involving 0.550 mol of an ideal gas. For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of A cyclic process. Figure 8.0 6.0 4.0 p(Pax 105) 2.0 a O 0.020 0.040 0.060 0.080 < 1 of 1 > V(m³) What was the temperature of this gas at points a, b, and c? Enter your answers separated by commas. ΤΥΠΙΑΣΦ 4 Ta Th. To 937.7,7654.12, 12296.59 Submit Part B X Incorrect; Try Again; 3 attempts remaining Submit Previous Answers Request Answer → How much work was done by or on the gas in this process? Express your answer in joules. W = 2.1x104 J Previous Answers Part C DMC ? ✔ Correct The work done in a process is the area under the curve in the pV diagram. The work done by the gas is positive since the volume increases. The magnitude of the work is the area under the curve: W = (2.0 × 105 Pa + 5.0 × 105 Pa) (0.070 m³ -0.010 m³) = 2.10 × 10¹ J Q= 3.4.104 How much heat had to be put in during the process to increase the internal energy of the gas by 1.00x104 J? Express your answer in joules. του 1920 K [w] ? Submit Previous Answers Request Answer J Review I Constants

The pV diagram in (Figure 1) shows a process abc involving 0.550 mol of an ideal gas. For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of A cyclic process. Figure 8.0 6.0 4.0 p(Pax 105) 2.0 a O 0.020 0.040 0.060 0.080 < 1 of 1 > V(m³) What was the temperature of this gas at points a, b, and c? Enter your answers separated by commas. ΤΥΠΙΑΣΦ 4 Ta Th. To 937.7,7654.12, 12296.59 Submit Part B X Incorrect; Try Again; 3 attempts remaining Submit Previous Answers Request Answer → How much work was done by or on the gas in this process? Express your answer in joules. W = 2.1x104 J Previous Answers Part C DMC ? ✔ Correct The work done in a process is the area under the curve in the pV diagram. The work done by the gas is positive since the volume increases. The magnitude of the work is the area under the curve: W = (2.0 × 105 Pa + 5.0 × 105 Pa) (0.070 m³ -0.010 m³) = 2.10 × 10¹ J Q= 3.4.104 How much heat had to be put in during the process to increase the internal energy of the gas by 1.00x104 J? Express your answer in joules. του 1920 K [w] ? Submit Previous Answers Request Answer J Review I Constants

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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